From the dimensional consideration, which of | vedclass

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Question

(a) By substituting the dimensions in $T = 2\pi \sqrt {\frac{{{R^3}}}{{GM}}} $
we get $\sqrt {\frac{{{L^3}}}{{{M^{ - 1}}{L^3}{T^{ - 2}} 	imes M}}} =

Vedclass · Accepted Answer

$T = 2\pi \sqrt {\frac{{{R^3}}}{{GM}}} $

Vedclass · Answer

(a) By substituting the dimensions in $T = 2\pi \sqrt {\frac{{{R^3}}}{{GM}}} $
we get $\sqrt {\frac{{{L^3}}}{{{M^{ - 1}}{L^3}{T^{ - 2}} 	imes M}}} = T$

From the dimensional consideration, which of the following equation is correct

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